【SDOI 2015】【BZOJ 3994】约数个数和

送个结论 d(n∗m)=∑d|n∑k|ne(gcd(d,k))
记 g(n)=∑ni=1⌊ni⌋
然后我们开始推
∑ni=1∑mj=1d(i∗j)=∑ni=1∑mj=1⌊ni⌋⌊mj⌋e(gcd(i,j))=∑ni=1∑mj=1⌊ni⌋⌊mj⌋∑d|i,d|jμ(d)=∑min(n,m)d=1μ(d)g(⌊nd⌋)g(⌊md⌋)
分块+预处理即可code：

#include
#include
#include
using namespace std;
int t,n,m,tot=0;
bool a[50001];
int prime[50001],mu[50001],s[50001],g[50001];
int calc(int maxn)
{
    int i,last,ans=0;
    for (i=1;i<=maxn;i=last+1)
      {
        last=maxn/(maxn/i);
        ans+=(last-i+1)*(maxn/i);
      }
    return ans;
}
long long work(int n,int m)
{
    int i,last;
    long long ans=0;
    if (n>m) swap(n,m);
    for (i=1;i<=n;i=last+1)
      {
        last=min(n/(n/i),m/(m/i));
        ans=ans+(long long)(s[last]-s[i-1])*(long long)(g[n/i])*(long long)(g[m/i]);      
      }
    return ans;
}
void prework(int maxn)
{
    int i,j;
    mu[1]=1;
    for (i=2;i<=maxn;++i)
      {
        if (!a[i])
          {prime[++tot]=i; mu[i]=-1;}
        for (j=1;j<=tot&&prime[j]*i<=maxn;++j)
          {
            a[prime[j]*i]=true;
            if (i%prime[j]==0)
              {
                mu[i*prime[j]]=0;
                break;  
              }
            mu[i*prime[j]]=-mu[i];
          }
      }
    for (i=1;i<=maxn;++i)
      s[i]=s[i-1]+mu[i];
    for (i=1;i<=maxn;++i)
      g[i]=calc(i);
}
int main()
{
    int i,minn;
    long long ans;
    prework(50000);
    scanf("%d",&t);
    while (t--)
      {
        scanf("%d%d",&n,&m);
        printf("%lld\n",work(n,m));
      }
}